Question: Let $h(x)=\sqrt{x}\ln(x)$. Find $h'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{2\sqrt{x}}\cdot \dfrac1x$ (Choice B) B $\dfrac{\sqrt{x}\ln(x)}{2}+\dfrac{1}{\sqrt{x}}$ (Choice C) C $\dfrac{1}{2\sqrt{x}}+ \dfrac1x$ (Choice D) D $\dfrac{\ln(x)}{2\sqrt{x}}+\dfrac{1}{\sqrt{x}}$
Answer: $h(x)$ is the product of two, more basic, expressions: $\sqrt{x}$ and $\ln(x)$. Therefore, the derivative of $h$ can be found using the product rule : $\begin{aligned} \dfrac{d}{dx}[u(x)v(x)]&=\dfrac{d}{dx}[u(x)]v(x)+u(x)\dfrac{d}{dx}[v(x)] \\\\ &=u'(x)v(x)+u(x)v'(x) \end{aligned}$ $\begin{aligned} &\phantom{=}h'(x) \\\\ &=\dfrac{d}{dx}\left(\sqrt{x}\ln(x)\right) \\\\ &=\dfrac{d}{dx}\left(\sqrt{x}\right)\ln(x)+\sqrt{x}\dfrac{d}{dx}(\ln(x))&&\gray{\text{The product rule}} \\\\ &=\dfrac{d}{dx}\left(x^{^{\frac{1}{2}}}\right)\ln(x)+\sqrt{x}\dfrac{d}{dx}(\ln(x))&&\gray{\text{Write }\sqrt{x} \text{ as a power}} \\\\ &=\dfrac12x^{^{-\frac{1}{2}}}\cdot \ln(x)+\sqrt{x} \cdot \dfrac1x&&\gray{\text{Differentiate }x^{^{\frac{1}{2}}}\text{ and }\ln(x)} \\\\ &=\dfrac{\ln(x)}{2\sqrt{x}}+\dfrac{\sqrt{x}}{x}&&\gray{\text{Simplify}} \\\\ &=\dfrac{\ln(x)}{2\sqrt{x}}+\dfrac{1}{\sqrt{x}}&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $h'(x)=\dfrac{\ln(x)}{2\sqrt{x}}+\dfrac{1}{\sqrt{x}}$